jungkook123
Badges: 6
Rep:
?
#1
Report Thread starter 4 weeks ago
#1
On a theme park, the carriage is fired vertically downwards from a heigh of 20m above ground level. The carriage is modelled as a particle and moves freely under gravity before the brakes are applied, 10m below ground level. When the brakes are applied, the carriage is travelling at 25ms-1.a) Find the speed of projection of the carriageb) Find the time taken for the carriage to reach ground levelI've been doing this question for ages and I have no clue on what to do, any help is appreciated
0
reply
mqb2766
Badges: 18
Rep:
?
#2
Report 4 weeks ago
#2
(Original post by jungkook123)
On a theme park, the carriage is fired vertically downwards from a heigh of 20m above ground level. The carriage is modelled as a particle and moves freely under gravity before the brakes are applied, 10m below ground level. When the brakes are applied, the carriage is travelling at 25ms-1.a) Find the speed of projection of the carriageb) Find the time taken for the carriage to reach ground levelI've been doing this question for ages and I have no clue on what to do, any help is appreciated
You must have done suvat?
Draw a diagram of the information in the problem and list which suvat variables are known and which you need to find. The suvat equation that you use should then link those variables.
0
reply
am3ricana
Badges: 5
Rep:
?
#3
Report 4 weeks ago
#3
(Original post by jungkook123)
On a theme park, the carriage is fired vertically downwards from a heigh of 20m above ground level. The carriage is modelled as a particle and moves freely under gravity before the brakes are applied, 10m below ground level. When the brakes are applied, the carriage is travelling at 25ms-1.a) Find the speed of projection of the carriageb) Find the time taken for the carriage to reach ground levelI've been doing this question for ages and I have no clue on what to do, any help is appreciated
Divide the question into two parts, the first 10, and the second 10m. Start with the first 10m and make your suvat list. Use the respective suvat formula to find u. do the same thing for the second 10m to find t.
0
reply
RogerOxon
Badges: 21
Rep:
?
#4
Report 4 weeks ago
#4
(Original post by am3ricana)
Divide the question into two parts, the first 10, and the second 10m. Start with the first 10m and make your suvat list. Use the respective suvat formula to find u. do the same thing for the second 10m to find t.
It's +20m and -10m.
0
reply
RogerOxon
Badges: 21
Rep:
?
#5
Report 4 weeks ago
#5
(Original post by jungkook123)
On a theme park, the carriage is fired vertically downwards from a heigh of 20m above ground level. The carriage is modelled as a particle and moves freely under gravity before the brakes are applied, 10m below ground level. When the brakes are applied, the carriage is travelling at 25ms-1.a) Find the speed of projection of the carriageb) Find the time taken for the carriage to reach ground levelI've been doing this question for ages and I have no clue on what to do, any help is appreciated
a: You have v, s and a, and want u. Which SUVAT equation works for that?
b: You now have u, a and s, and want t. Which SUVAT equation works for that?

Be careful to apply your convention for positive direction consistently.

You will get two solutions for a, but know that it is projected downwards, so can discount one.
Last edited by RogerOxon; 4 weeks ago
0
reply
jungkook123
Badges: 6
Rep:
?
#6
Report Thread starter 4 weeks ago
#6
(Original post by RogerOxon)
a: You have v, s and a, and want u. Which SUVAT equation works for that?
b: You now have u, a and s, and want t. Which SUVAT equation works for that?

Be careful to apply your convention for positive direction consistently.

You will get two solutions for a, but know that it is projected downwards, so can discount one.
what would v be in question a? 0?
0
reply
jungkook123
Badges: 6
Rep:
?
#7
Report Thread starter 4 weeks ago
#7
(Original post by mqb2766)
You must have done suvat?
Draw a diagram of the information in the problem and list which suvat variables are known and which you need to find. The suvat equation that you use should then link those variables.
I did do that but I keep getting the wrong answer
0
reply
RogerOxon
Badges: 21
Rep:
?
#8
Report 4 weeks ago
#8
(Original post by jungkook123)
what would v be in question a? 0?
For a:

v is the final speed, at s=30m (+20 to -10), and is 25m/s. a is the acceleration due to gravity, which is ~10m/s^2 (I don't know what you were told to take this as). I'm taking down as the positive direction, with s=0 being the initial point, so that all my numbers should be positive.

Hint: For a, use v^2=u^2+2as and solve for u.
0
reply
RogerOxon
Badges: 21
Rep:
?
#9
Report 4 weeks ago
#9
(Original post by jungkook123)
I did do that but I keep getting the wrong answer
Please post your working.
0
reply
jungkook123
Badges: 6
Rep:
?
#10
Report Thread starter 4 weeks ago
#10
(Original post by RogerOxon)
For a:

v is the final speed, at s=30m (+20 to -10), and is 25m/s. a is the acceleration due to gravity, which is ~10m/s^2 (I don't know what you were told to take this as). I'm taking down as the positive direction, with s=0 being the initial point, so that all my numbers should be positive.

Hint: For a, use v^2=u^2+2as and solve for u.
I finally got the answer, thank you sm!!!!!
0
reply
RogerOxon
Badges: 21
Rep:
?
#11
Report 4 weeks ago
#11
(Original post by jungkook123)
I finally got the answer, thank you sm!!!!!
Good. Did you see that you get two answers for u? Do you understand why?

Have you done part b?
0
reply
jungkook123
Badges: 6
Rep:
?
#12
Report Thread starter 4 weeks ago
#12
(Original post by RogerOxon)
Good. Did you see that you get two answers for u? Do you understand why?

Have you done part b?
no I only got one answer for u which was 6.08, I haven't done part b yet
0
reply
RogerOxon
Badges: 21
Rep:
?
#13
Report 4 weeks ago
#13
(Original post by jungkook123)
no I only got one answer for u which was 6.08, I haven't done part b yet
Your answer is correct, so I'll post the working:

v^2=u^2-2as

\therefore u^2=v^2-2as={25}^2-2*9.80665*30=36.601

\therefore u=\pm 6.05 m{s}^{-1} (3 s.f.)

We're told that the carriage is projected downwards, so can ignore the negative result. It's useful to understand it though. If you throw something upwards, with only gravity acting upon it, it later will pass through the initial point with the same speed, but in the other direction. The negative result here is telling you that, if you projected the carriage upwards at the same speed, it would later pass the initial point at the same speed downwards, so either initial speed is a solution to the equation.
0
reply
jungkook123
Badges: 6
Rep:
?
#14
Report Thread starter 4 weeks ago
#14
(Original post by RogerOxon)
Your answer is correct, so I'll post the working:

v^2=u^2-2as

\therefore u^2=v^2-2as={25}^2-2*9.80665*30=36.601

\therefore u=\pm 6.05 m{s}^{-1} (3 s.f.)

We're told that the carriage is projected downwards, so can ignore the negative result. It's useful to understand it though. If you throw something upwards, with only gravity acting upon it, it later will pass through the initial point with the same speed, but in the other direction. The negative result here is telling you that, if you projected the carriage upwards at the same speed, it would later pass the initial point at the same speed downwards, so either initial speed is a solution to the equation.
you were a big help, thank u sm again
0
reply
RogerOxon
Badges: 21
Rep:
?
#15
Report 4 weeks ago
#15
(Original post by jungkook123)
I haven't done part b yet
We now have the initial speed, u, from part a. We know s (distance from the initial point to the ground) and a. We don't know v (the speed at ground level), You need to choose a SUVAT equation that has only s, u, a and t, then solve for t.

Again, you will get two solutions, of which one will be negative. If you draw a graph of displacement verses time, you will see why. For the maths t=0 isn't some special value, but it is practically, as the carriage wasn't moving before it.
0
reply
jungkook123
Badges: 6
Rep:
?
#16
Report Thread starter 4 weeks ago
#16
(Original post by RogerOxon)
We now have the initial speed, u, from part a. We know s (distance from the initial point to the ground) and a. We don't know v (the speed at ground level), You need to choose a SUVAT equation that has only s, u, a and t, then solve for t.

Again, you will get two solutions, of which one will be negative. If you draw a graph of displacement verses time, you will see why. For the maths t=0 isn't some special value, but it is practically, as the carriage wasn't moving before it.
I got the answer t=1.49(3.sf) and t=-2.73(3.sf), considering time can't be negative I got t=1.49s(3.sf)?
0
reply
RogerOxon
Badges: 21
Rep:
?
#17
Report 4 weeks ago
#17
(Original post by jungkook123)
I got the answer t=1.49(3.sf) and t=-2.73(3.sf), considering time can't be negative I got t=1.49s(3.sf)?
I haven't done part b. A quick check is to use v=u+at, which gives ~21 m/s. Now use s=\frac{u+v}{2}t. That gives s as ~20m, which matches, confirming your answer. It's always good to check your answers, especially when it can be done quickly.
Last edited by RogerOxon; 4 weeks ago
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

What are you most likely to do if you don't get the grades you were expecting?

Go through Clearing (210)
38.53%
Take autumn exams (164)
30.09%
Look for a job (19)
3.49%
Consider an apprenticeship (22)
4.04%
Take a year out (97)
17.8%
Something else (let us know in the thread!) (33)
6.06%

Watched Threads

View All